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## Q. 2.1

The floor of a classroom is supported by the bar joists shown in Fig. 2–15a. Each joist is 15 ft long and they are spaced 2.5 ft on centers. The floor itself is made from lightweight concrete that is 4 in. thick. Neglect the weight of the joists and the corrugated metal deck, and determine the load that acts along each joist.

## Verified Solution

The dead load on the floor is due to the weight of the concrete slab. From
Table 1–3

 Walls psf ${kN}/{m^{2} }$ 4-in. (102 mm) clay brick 39 1.87 8-in. (203 mm) clay brick 79 3.78 12-in. (305 mm) clay brick 115 5.51 Frame Partitions and Walls Exterior stud walls with brick veneer 48 2.30 Windows, glass, frame and sash 8 0.38 Wood studs $\left(2\times 4\right) in.$,$\left(51\times 102 mm\right)$ unplastered 4 0.19 Wood studs $\left(2\times 4\right) in.$,$\left(51\times 102 mm\right)$ plastered one side 12 0.57 Wood studs $\left(2\times 4\right) in.$,$\left(51\times 102 mm\right)$ plastered two sides 20 0.96 Floor Fill Cinder concrete, per inch (mm) 9 0.017 Lightweight concrete, plain, per inch (mm) 8 0.015 Stone concrete, per inch (mm) 12 0.023 Ceilings Acoustical fiberboard 1 0.05 Plaster on tile or concrete 5 0.24 Suspended metal lath and gypsum plaster 10 0.48 Asphalt shingles 2 0.10 Fiberboard, $\frac{1}{2} in.$ (13 mm) 0.75 0.04

for 4 in. of lightweight concrete it is $\left(4\right)\left(8 {lb}/{ft^{2} }\right)=32 {lb}/{ft^{2} }$

From Table 1–4, the live load for a classroom is $40 {lb}/{ft^{2} }$

 TABLE 1-4   .  Minimum Live Loads Live Load Live Load Occupancy or Use psf ${kN}/{m^{2}}$ Occupancy or Use psf ${kN}/{m^{2}}$ Assembly areas and theaters Residential Fixed seats 60 2.87 Dwellings (one- and two-family) 40 1.92 Movable seats 100 4.79 Hotels and multifamily houses Dance halls and ballrooms 100 4.79 Private rooms and corridors 40 1.92 Garages (passenger cars only) 50 2.40 Public rooms and corridors 100 4.79 Office buildings Schools Lobbies 100 4.79 Classrooms 40 1.92 Offices 50 2.40 Corridors above first floor 80 3.83 Storage warehouse Light 125 6.00 Heavy 250 11.97

Thus the total floor load is $32 {lb}/{ft^{2} }+ 40{lb}/{ft^{2} }=72 {lb}/{ft^{2} }$

For the floor system, $L_{1}=2.5 ft$ and $L_{2}=15 ft$.  Since ${L_{2}}/{L_{1}}\gt1.5$ the concrete slab is treated as a one-way slab. The tributary area for each joist is shown in Fig. 2–15b. Therefore the uniform load along its length is

$w=72 {lb}/{ft^{2} }\left(2.5 ft\right)=180 {lb}/{ft}$

This loading and the end reactions on each joist are shown in Fig. 2–15c.